


im 



Hollinger Corp. 
pH 8.5 



LC 6301 
.U5 P55 
Copy 1 






j/ 



ZlniversitY and School Extensien 



PLANE TRIGONOMETRY. 



A. W. PHILLIPS, 

1889. 

Yale University. 



le g 



^ 01 



\ 



Copyright, 

1889, 

By A. W. PHILLIPS. 



V 



Press of J. J. Little & Co. : 

Astor Place, New York. 



Course in Trigonometry. 



OBJECT OF THE COURSE. 

§ i. The Plane Trigonometry is intended to teach the student 
the derivation of the formulas and the methods of solving, both 
with and without the use of Logarithms, all the cases of right- 
angled and oblique-angled triangles, together with the application 
of these methods to the measurement of heights and distances, 
to field surveying with the compass and theodolite, and also to the 
simpler problems of" Navigation. 

The Spherical Trigonometry covers the derivation of the 
formulas and the methods of solving all the cases of right-angled 
and oblique-angled spherical triangles, with their applications to the 
simple problems of the terrestrial and celestial globe. 

METHODS OF STUDY. 

§ 2. The student may make use of any of the standard element- 
ary text-books used, in the academies and colleges, and can master 
topic after topic in the order indicated in the syllabus, working the 
problems from the text-book. 

It is strongly recommended, where it is possible, to use the 
compass and theodolite or even more roughly constructed instru- 
ments for measuring angles, and to work problems constructed from 
the student's own measurements. 



UNIVERSITY EXTENSION. 



THE DEFINITIONS OF THE TRIGONOMETRIC 
FUNCTIONS. 

§ 3. The definitions of the Trigonometric functions, viz.: sines, 
cosines, tangents, etc., are given either as certain lines drawn in or 
about a circle of unit radius, or as certain ratios of the sides of a 
so-called triangle of reference. 

The advantage of the first method is that the student is always 
able to represent these functions by definite lines in a figure, which 
gives his problems more of a reality than the second method. 

In the method of ratios which is more generally adopted in the 
recent works published in this country, the early conceptions are 
much more difficult ; but since these definitions are the formulas 
of right triangles the student is greatly aided in becoming familiar 
with the fundamental principles in the solution of triangles. 

The following figures will furnish an example of each of the 
several functions in the four quadrants : 

A' JL' 





Fig. 1. 
1st Quadrant. 



Fig. 2. 
2d Quadrant. 



COURSE IN TRIGONOMETRY. 




Fig. 3. 
3d Quadrant. 




Fig. 4. 

4th Quadrant. 



For the angle AOBor the arc A B in each figure. 
B E is the sine, O D is the secant, 

O E the cosine, O C the cosecant, 

A D the tangent. A' C the cotangent. 



RULE OF SIGNS. 

For all -vertical lines, sines and tangents : 

Those above the horizontal diameters are + . 
Those below, — . 

For all horizontal lines, cosines and cotangents : 

Those on the right of the vertical diameter are + . 
Those on the left, — . 

For oblique lines, secant and cosecant : 

Those reckoned from the centre towards the extremity of 
the arc are + . 



UNIVERSITY EXTENSION. 



Those reckoned from the centre in the direction opposite 
the extremity of the arc are — . 
[The radius is always to be regarded as positive.] 

TABLE SHOWING THE SIGNS IN EACH QUADRANT. 





ISt. 


2d. • 


3d- 


4th. 


Sine 


+ 


+ 


- 


- 


Cosine 


+ 


- 


- 


+ 


Tangent 


+ 


- 


+ 


- 


Cotangent .... 


+ 


- 


+ 


- 


Secant 


+ 


- 


- 


+ 


Cosecant 


+ 


+ 


- 





After learning the definitions of the Trigonometric functions, 
and the signs of these functions in the several quadrants, the next 
step is to derive the following 



FUNDAMENTAL FORMULAS 

FROM A FIGURE. 



(i) sec Oi 



COS X 



(2) cosec x == 



sin x 



COURSE IN TRIGONOMETRY. 



Sill X COS X 

(3) tan x — (4) cot x = 

cos x sin x 

1 

(5) tan x — (6) sin 2 x + cos 2 x — 1. 

cot x 

(7) sec 2 x = 1 + tan 2 x. (8) cosec 2 .# = 1 + cot 2 x. 



VALUES OF FUNCTIONS OF CERTAIN ANGLES 

§ 5. VERY FREQUENTLY USED. 



sin == 


cos = 1. 


sin 30 = 


1 iTj 




, cos 30 — 

2 2 


sin 45 = 


V~T VJ 


, cos 45 — 

2 2 


sin 6o° = 


VJ 1 

/■ _ 





1. Draw figures and show < sin 45 



2 2 

sin 90 = 1 ; cos 90 — o. 

2. From these values of the sin and cos, derive values of the tan, 

cot, sec, cosec of o°, 30 , 45 , 6o°, 90 . 

3. From these values derive, by the aid of a figure, the sin, cos, tan, 

cot, sec, and cosec, of 120 , 135°, 150 , 180 , 210 , 225 , 
240 , 270 , 300 , 330 , 360 . 



UNIVERSITY EXTENSION. 



GRAPHICAL SOLUTION OF TRIANGLES. 

§ 6. The parts of the triangle which are given may be laid off 
—the sides on any convenient scale, and the angles by means of a 
protractor — and the remaining sides and angles may then be 
measured. The results will give a good approximation if the meas- 
urements are carefully made — and will serve also as a check on the 
calculations. 

SOLUTION OF RIGHT TRIANGLES. 

§ 7. Let A, B, C be a right triangle. A the right angle, and B 
the acute angle at the base. Call the sides opposite A, B, and C 
respectively #, b, c. 

b opposite side 

sin B = — or sin of an angle = 

a hypotenuse 

c adjacent side 

cos B = — or cos of an angle — — 

a hypotenuse 

b opposite side 

tan B = — or tan of an angle — 

c adjacent side 

#2 = ^2_|-^2 or square of hypotenuse = sum of squares 
of the other two sides. 



COURSE IN TRIGONOMETRY. 



APPLICATIONS. 

§ 8. To problems in heights and distances where the triangles 
formed are right-angled. 

To problems in field surveying with the compass. 

To problems in Navigation in Plane Sailing, Middle Latitude 
Sailing, and Mercator's Sailing. 



EXAMPLES. 

Find the remaining parts in each of the following right trian- 
gles. Given : 

Base 60 feet, angle at the base 43 °. 

Base 75 feet, angle at vertex 76 18' 12". 

Perpendicular 62 feet, angle at the base 47 18'. 

Hypotenuse 87 feet, angle at vertex 8$° 10'. 

Hypotenuse 50 feet, base 37 ft. 

Find the distance across a stream, or to a given inaccessible 
object, by constructing a right-angled triangle and measuring 
one side and one angle. 

7. Find the height of a tree, a steeple, or any object, by measur- 
ing the base of a triangle, and the angle of elevation of the 
top of the tree. 



UNIVERSITY EXTENSION. 



8. Find the contents of a field by dividing it up into right-angled 

triangles and measuring such portions as are necessary in 
order to find the area. 

9. Survey a field with a compass and chain, and explain the 

method of finding its contents. 

10. In Plane Sailing, show how to find : 

(a) The departure, given the bearing and distance. 

(b) The difference of latitude, given the departure and dis- 

tance. 

(c) The distance, given the beaming and difference of latitude. 

[Draw a triangle and label the sides and the angle properly.] 

11. In Parallel Sailing, draw a right triangle and label its parts so 

as to show the relation between the distance measured on a 
parallel, the latitude of the parallel, and the radius of the 
earth. 

12. In Mercators Sailing, draw a right triangle and label it so as 

to show the relation between the bearing, the distance sailed, 
the departure, the difference of latitude, the difference of 
longitude, and the meridianal parts. 

13. Given the latitude and longitude of two places, find the bearing, 

the departure, and the distance sailed in going from one to 
the other. 



COURSE IN TRIGONOMETRY. 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES 

WITHOUT LOGARITHMS. 

§ 9. Let A, B, C be an oblique-angled triangle, of which A B is 
the base, and the angle B either acute or obtuse. The sides opposite 
A, B, and C are respectively a, b, and c. Let fall the perpendicular 
C D on A B or A B produced. 

a sin A 

Then b sin A = a sin B ; or — = (A.) 

b sin B 

Making in turn a and b the base, we obtain similar expressions 

b c 

for — and — . We therefore have the Rule: The sides of a triangle 

c a 

are proportional to the sines of the opposite angles. 

Again, we have from the triangle a 2 = c 2 + b 2 — 2 be cos A (B.) 

Making in turn a and b the base, we obtain similar expressions 
for b 2 and c 2 . We therefore have the Rule : The square of the 
side opposite any angle of a triangle is equal to the sum of the squares 
of the other two sides diminished by twice the product of these two sides 
into the cosine of the opposite angle. 



CASES. 

Two sides and an opposite angle 
Two angles and an opposite side 

Two sides and included angle 
Three sides 



Formula (A). 
Formula (B). 



12 



UNIVERSITY EXTENSION. 



DEVELOPMENT OF FORMULAS. 
§ 10. The formulas for the sine and cosine of the sum and dif- 
ference of two angles may be derived directly from the following 
figures: 





In Fig. 5, call the angle A O B or arc A B x, and B O C or B C 
y ; then A O C or A C is x + y. O K is the cos y ; C K is perpen- 
dicular to O B, and is the sin 7 ; C D and K H are perpendicular to 
O A ; K L is perpendicular to C D. The angle KCL = AOB = x 

CD^KH+CL 

C L = t) K 90s x; = sin y cos x. 

K H = O K sin x = sin x cos y. 
C D = sin (x + y) = sin x cosy + cos x smy. 



O D = O H - L K. 

O H = O K cos x = cos x cosy. 

L K = CK sin y = sin x sin y. 

O D — cos (x + y) = cos x cosy — sin x siny. 



COURSE IN TRIGONOMETRY. 1 3 

In Fig. 6, call angle A O B or arc A .B x, and COBorCB;. 
The angle CHF = i,CHis the sin of y, and O H the cosine. H D 
'is perpendicular to O A, and C F to H D. 

CE = HD-HF. 

H D = O H sin x = sin x cos y. 

HF =HC cos x = cos x sin y. 

C E = sin (x — y) — sin x cos y — cos x sin y. 

OE = OD + FC. 

OD = OH cos x = cos x cos y. 

F C = H C sin y = sin x siny. 

OE = cos (x — y) = cos x cosy -f sin x siny. 

Collecting these formulas together, we have : 
Sin (x + y) = sin x cosy + cos x siny. (i.) 
Sin (x — y) = sin x cos y — cos x sin y. (2.) 
Cos {x + y) = cos^: cosjy — sin x siny. (3.) 
Cos (x — y) = cos x cosy 4- sin jp sin jy. (4.) 

Putting j = x equation (1) becomes : 
Sin 2 x = 2 sin x cos x. (5.) 

Equation (3) also becomes : 

Cos 2 x = Cos 2 .* — sin 2 #, or cos 2^ = 2 cos 2 x — 1, or 
cos 2 .# = 1 — 2 sin 2 .#. (6.) 

Adding (1) and (2) : 

Sin (x + y) + sin (.* — jy) = 2 sin 3; cos>». (7.) 



14 UNIVERSITY EXTENSION. 

Subtracting (2) from (1) : 

Sin (x + y) — sin (x — y) = 2 cos x sin y. (8.) 

Adding (3) and (4) : 

Cos (x + y) + cos (x — ■ y) = 2 cos x cosy. (9.) 

Subtracting (3) from (4) : 

Gos (x — y) — cos (x -\- y) = 2 sin x sin y. (10.) 

Putting x -f- y = A, and o: — y = E, (7), (8), (9), (10), become 

A + B A - B. 

Sin A + sin B = 2 sin cos (n-) 

2 2 

A -f B A - B. 

Sin A — sin B = 2 sin cos ( I2 -) 

2 2 

A + B A - B. 

Cos A + cos B = 2 cos ! cos (13.) 

A + B A - B. 
Cos B — cos A = 2 sin sin ( I 4-) 



EXERCISES. 



-n • t/ A / 1 cos x 

i. rrove sm % x = A / 



x=j/i. 



T/ x + COS X 

2. rrove cos V 



COURSE IN TRIGONOMETRY 



3. Derive sin 3 x 

4. Derive cos 3 ^ 

tan x + tan y 

5. Prove tan (x + y) = 

1 — tan x tan y 

tan jc — tan y 
1 4- tan x tan y 
cot a- cot y — 1 

cot y + cot x 
cot x cot y + 1 

cot y — cot x 



6. Prove tan (x — y) = 

7. Prove cot (x + y) = 

8. Prove cot (a* — ^) = 



9. Show that Formulas (1), (2), (3), (4) are true when either x or 
y or both are greater than 90 

10. Show that sin (90 + x) = cos x 

11. Show that cos (90 + X) = sin .r 

12. Show that tan (90 + x) = cot x 

13. Derive sine and cosine of 15 , 75 



-Y*r 



14. Show sine of 18 

15. Given sin x + cos x — 1 : find x 



l6 UNIVERSITY EXTENSION. 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES 

BY THE USE OF LOGARITHMS. 

§ ii. The tables of logarithmic sines, cosines, tangents, and 
cotangents are the logarithms of the numbers in the natural tables of 
logarithmic functions — the characteristic in each case being increased 
by 10, to avoid the use of negative characteristics. 

The text-books show how to use these logarithms. 

Since it would be a great inconvenience to use logarithms in 
formulas where the terms are connected by the signs plus and minus, 
it is necessary, in order to adapt formulas (A) and (B), oblique- 
angled triangles, to logarithmic computation, to transform them so 
the terms shall be products or quotients. 

The foregoing chapter in Trigonometric analysis furnishes the 
means for making these transformations, and also forms the basis 
for solving many problems not necessary in the solution of triangles. 

a sin A 

Formula (A) : — = by composition and division becomes 

b sin B 

a 4- b sin A + sin B 



sin A — sin B 



Dividing (n) by (12) and reducing: 
sin A + sin B tan X A (A + B) 



sin A — sin B tan X A (A — B) 



COURSE IN TRIGONOMETRY. 1 7 

Whence 

a + b tan % (A + B) 

- (C) 

a - b tan ^ (A — B) 

Formula (B) : a 2 = e 2 + b 2 — 2 be cos A, may be written : 
cos A = 



2 be 
From (6) : 

e 2 + b 2 - a 2 (£ + c + a) (b + c - a) 
2 cos 2 y 2 A= 4- 1 = 



2 &■ 2 be 

(IS-) 
From (6) : 

c 2 + £ 2 - a 2 (0— * + f) (0 + .4 - * ) 

2 sin 2 ^ A == 1 — ■ = 

2 be 2 be 

(.6.) 
# -f b + <: 

Putting -= s, and dividing (16) by (15) and extracting 

2 
the square root : 



s (s—a 



(s-a) (D.) 



SUMMARY. 



a sin A 

12. - = (A.) 

b sin B 



UNIVERSITY EXTENSION. 



a + b tan y 2 (A + B) 

= (C) 

a - b tan ^ (A - B) 



a / {s - c) (s - b) 
tan y 2 A = \/ -± M ^ (D.) 

A + B + C = 180 . (E.) 



CASES. — Formulas for complete solution. 

Two sides and opposite angle. (A), (E). 

Two angles and opposite side. (A), (E). 

Two sides and included angle. (E), (C), (A). 

Two angles and included side. (E), (A). 

Three sides. (D), (A), (E). 



In the above formulas and also in the . 

cY A 
formulas of oblique-angled spherical triangles, ^ a 

the formulas will hold true if, in place of each rk 

letter, we substitute the one next in order, taken c b 

C" B 

in the direction of the arrow in the figure 

annexed. 

This amounts to the same thing as turning the triangle, so as 

to make for the base the side next to the base in order. This 

will give for each formula a group of three formulas. 



LIBRARY OF CONGRESS 



029 944 930 



LIBRARY OF CONGRESS 



029 944 930 



